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Question: Prove that \(\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta...

Prove that tan⁡(π4+θ)−tan⁡(π4−θ)=2tan⁡2θ\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = 2\tan 2\theta

Explanation

Solution

Hint : We will solve this problem by first considering the L.H.S. of the given equation. Then we will consider the formulas of tan and expand the terms. And after solving and simplifying we will get to the answer.

Complete step-by-step answer :
First, we will take L.H.S and solve
=tan⁡(π4+θ)−tan⁡(π4−θ)= \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right)
Now by using the formulas tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}and tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}, we will expand the terms of L.H.S.
=tan⁡π4+tan⁡θ1−tan⁡π4tan⁡θ−tan⁡π4−tan⁡θ1+tan⁡π4tan⁡θ= \dfrac{{\tan \dfrac{\pi }{4} + \tan \theta }}{{1 - \tan \dfrac{\pi }{4}\tan \theta }} - \dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan \dfrac{\pi }{4}\tan \theta }}
Value of tan⁡π4\tan \dfrac{\pi }{4}is 1. So, replacing it in the above equation.
=1+tan⁡θ1−tan⁡θ−1−tan⁡θ1+tan⁡θ= \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} - \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}
Now, we will take the LCM of denominators. And by using the formula a2 - b2 = (a - b) (a + b), we will replace the values in the denominator.
=(1+tan⁡θ)2−(1−tan⁡θ)212−tan⁡2θ= \dfrac{{{{(1 + \tan \theta )}^2} - {{(1 - \tan \theta )}^2}}}{{{1^2} - {{\tan }^2}\theta }}
Now, by using formulas (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2aband (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, we will expand the terms of numerator.
=12+tan⁡2θ+2tan⁡θ−(12+tan⁡2θ−2tan⁡θ)1−tan⁡2θ= \dfrac{{{1^2} + {{\tan }^2}\theta + 2\tan \theta - \left( {{1^2} + {{\tan }^2}\theta - 2\tan \theta } \right)}}{{1 - {{\tan }^2}\theta }}
=1+tan⁡2θ+2tan⁡θ−1−tan⁡2θ+2tan⁡θ1−tan⁡2θ= \dfrac{{1 + {{\tan }^2}\theta + 2\tan \theta - 1 - {{\tan }^2}\theta + 2\tan \theta }}{{1 - {{\tan }^2}\theta }}
=4tan⁡θ1−tan⁡2θ= \dfrac{{4\tan \theta }}{{1 - {{\tan }^2}\theta }}
Now, we can rewrite this equation as
=2×(2tan⁡θ)1+tan⁡2θ= \dfrac{{2 \times \left( {2\tan \theta } \right)}}{{1 + {{\tan }^2}\theta }}
=2×(tan⁡θ+tan⁡θ)1−tan⁡θtan⁡θ= \dfrac{{2 \times \left( {\tan \theta + \tan \theta } \right)}}{{1 - \tan \theta \tan \theta }}
=2×(tan⁡θ+tan⁡θ)1−tan⁡θtan⁡θ= 2 \times \dfrac{{\left( {\tan \theta + \tan \theta } \right)}}{{1 - \tan \theta \tan \theta }}
By using the formula tan (A+B). we can rewrite the equation as:
=2×tan⁡(θ+θ)= 2 \times \tan \left( {\theta + \theta } \right)
=2tan⁡2θ= 2\tan 2\theta
This is equal to RHS.
Hence, proved.

Note : In these types of questions we have to know the general formula of trigonometry. Students should remember trigonometric identities and important formulas for solving these types of problems.
Also, Simplify the equations to the standard formula to get the desired answer.
There is also an alternative method to solve this question. we can first convert the terms which are in tan to sine or cosine. Then, similarly take the L.H.S side to solve and use these kinds of formulas which we used above to solve the problem. And again, at the end change the sine or cosine terms in tan to get the equation proved.